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Is W a subspace of the vector space?

  1. Oct 24, 2007 #1
    W={(x1,x2,x3):x[tex]^{2}_{1}[/tex]+x[tex]^{2}_{2}[/tex]+x[tex]^{2}_{3}[/tex]=0} , V=R^3

    Is W a subspace of the vector space?
    from what i understand for subspace to be a subspace it has to have two conditions:
    1.must be closed under addition
    2.must be closed under multiplication

    so....
    I pick a vector s=(s1,s2,s3) and a second vector t=(t1,t2,t3).

    for the addition i get:
    s+t=(s1+t1,s2+t2,s3+t3)//so its closed under addition

    for multiplication i get:
    cs=c(s1,s2,s3)=(cs1,cs2,cs3)//closed under multiplicartion

    what i dont understand is how the condition x[tex]^{2}_{1}[/tex]+x[tex]^{2}_{2}[/tex]+x[tex]^{2}_{3}[/tex]=0 would come into play. how do i use this condition in this problem?
     
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2

    Dick

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    You mean V=R^3, right? And your 'proof' just proves R^3 is a vector space. It doesn't say anything about W. What is the set W? Can you find a point in it? Can you describe a general point in W?
     
  4. Oct 24, 2007 #3

    radou

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    Of which vector space? You should be more precise, although it is obvious you meant R^3.

    It is more practical to express this as one single condition; W is a subspace of V if and only if for any two vectors x, y from W, and any scalars a, b, ax + by is in W.

    So, take any two vectors from W, let's say x = (x1, x2, x3) and y = (y1, y2, y3). Now write the linear combination ax + by, and see if the components of ax + by satisfy the condition for a vector to be in W. Also, while doing this, keep in mind that the components of x and y do satisfy this very condition!
     
  5. Oct 24, 2007 #4
    my book says this:

    If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following closure conditions hold.
    1.u and v are in W, then u+v is in w.
    2if u is in W and c is a scalar, then cu is in W.
     
  6. Oct 24, 2007 #5

    Hurkyl

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    So what does "x is in W" mean here?
     
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